17. Large Cardinals - Delft University of Technology17. Large Cardinals 287 Now let κ be a measurable cardinal, and let U be a nonprincipal κ-com- plete ultraﬁlter on κ.Letd (the

17. Large Cardinals

The theory of large cardinals plays central role in modern set theory. Inthis chapter we begin a systematic study of large cardinals. In addition tocombinatorial methods, the proofs use techniques from model theory.

Ultrapowers and Elementary Embeddings

We start with the following theorem that introduced the technique of ultra-powers to the study of large cardinals.

Theorem 17.1 (Scott). If there is a measurable cardinal then V �= L.

Ultrapowers were introduced in Chapter 12. We now generalize the tech-nique to construct ultrapowers of the universe. Let U be an ultrafilter ona set S and consider the class of all functions with domain S. Following(12.3) and (12.4) we define

f =∗ g if and only if {x ∈ S : f(x) = g(x)} ∈ U,

f ∈∗ g if and only if {x ∈ S : f(x) ∈ g(x)} ∈ U.

For each f , we denote [f ] the equivalence class of f in =∗ (recall (6.4)):

[f ] = {g : f =∗ g and ∀h (h =∗ f → rank g ≤ rankh)}.

We also use the notation [f ] ∈∗ [g] when f ∈∗ g.Let Ult = UltU (V ) be the class of all [f ], where f is a function on S, and

consider the model Ult = (Ult,∈∗). �Los’s Theorem 12.3 holds in the presentcontext as well: If ϕ(x1, . . . , xn) is a formula of set theory, then

Ult � ϕ([f1], . . . , [fn]) if and only if {x ∈ S : ϕ(f1(x), . . . , fn(x))} ∈ U.

If σ is a sentence, then Ult � σ if and only if σ holds; the ultrapoweris elementarily equivalent to the universe (V,∈). The constant functions ca

are defined, for every set a, by (12.12), and the function j = jU : V → Ult,defined by jU (a) = [ca] is an elementary embedding of V in Ult:

(17.1) ϕ(a1, . . . , an) if and only if Ult � ϕ(ja1, . . . , jan)

whenever ϕ(x1, . . . , xn) is a formula of set theory.

286 Part II. Advanced Set Theory

The most important application of ultrapowers in set theory are those inwhich (Ult,∈∗) is well-founded. As we show below, well-founded ultrapowersare closely related to measurable cardinals.

The model UltU (V ) is well-founded if (i) every nonempty set X ⊂ Ulthas a ∈∗-minimal element, and (ii) ext(f) is a set for every f , where

ext(f) = {[g] : g ∈∗ f}.

The second condition is clearly satisfied for any ultrafilter U : For every g ∈∗ fthere is some h =∗ g such that rankh ≤ rank f . As for the condition (i), thisis satisfied if and only if there exists no infinite descending ∈∗-sequence

f0 �∗ f1 �∗ . . . �∗ fk �∗ . . . (k ∈ ω)

of elements of the ultrapower.

Lemma 17.2. If U is a σ-complete ultrafilter, then (Ult,∈∗) is a well-founded model.

Proof. We shall show that there is no infinite descending ∈∗-sequence in Ultif U is a σ-complete ultrafilter on S. Let us assume that f0, f1, . . . , fn, . . .is such a descending sequence. Thus for each n, the set

Xn = {x ∈ S : fn+1(x) ∈ fn(x)}

is in the ultrafilter. Since U is σ-complete, the intersection X =⋂∞

n=0 Xn isalso in U and hence nonempty; let x be an arbitrary element of X . Then wehave

f0(x) � f1(x) � f2(x) � . . .

an infinite descending �-sequence, which is a contradiction. ��

By the Mostowski Collapsing Theorem every well-founded model is iso-morphic to a transitive model. Thus if U is σ-complete, there exists a one-to-one mapping π of Ult onto a transitive class such that f ∈∗ g if and onlyπ([f ]) ∈ π([g]). In order to simplify notation, we shall identify each [f ] withits image π([f ]). Thus if U is σ-complete, the symbol Ult denotes the transi-tive collapse of the ultrapower, and for each function f on S, [f ] is an elementof the transitive class Ult; we say the function f represents [f ] ∈ Ult.

Thus if U is a σ-complete ultrafilter, M = UltU (V ) is an inner model andj = jU is an elementary embedding j : V → M .

If α is an ordinal, then since j is elementary, j(α) is an ordinal; moreover,α < β implies j(α) < j(β). Thus we have α ≤ j(α) for every ordinal num-ber α. Note that j(α + 1) = j(α)+1, and j(n) = n for all natural numbers n.It is also easy to see that j(ω) = ω: If [f ] < ω, then f(x) < ω for almostall x ∈ S, and by σ-completeness, there exists n < ω such that f(x) = n foralmost all x. By the same argument, if U is λ-complete, then j(γ) = γ for allγ < λ.

17. Large Cardinals 287

Now let κ be a measurable cardinal, and let U be a nonprincipal κ-com-plete ultrafilter on κ. Let d (the diagonal function) be the function on κdefined by

d(α) = α (α < κ).

Since U is κ-complete every bounded subset of κ has measure 0 and so forevery γ < κ, we have d(α) > γ for almost all α. Hence [d] > γ for all γ < κand thus [d] ≥ κ. However, we clearly have [d] < j(κ) and it follows thatj(κ) > κ.

We have thus proved that if there is a measurable cardinal, then thereis an elementary embedding j of the universe in a transitive model M suchthat j is not the identity mapping; j is a nontrivial elementary embedding ofthe universe.

Proof of Theorem 17.1. Let us assume V = L and that measurable cardinalsexist; let κ be the least measurable cardinal. Let U be a nonprincipal κ-complete ultrafilter on κ and let j : V → M be the corresponding elementaryembedding. As we have shown, j(κ) > κ.

Since V = L, the only transitive model containing all ordinals is theuniverse itself: V = M = L. Since j is an elementary embedding and κ is theleast measurable cardinal, we have

M � j(κ) is the least measurable cardinal;

and hence, j(κ) is the least measurable cardinal. This is a contradiction sincej(κ) > κ. ��

If there exists a measurable cardinal, then there exists a nontrivial elemen-tary embedding of the universe. Let us show that conversely, if j : V → M isa nontrivial elementary embedding then there exists a measurable cardinal.

Lemma 17.3. If j is a nontrivial elementary embedding of the universe, thenthere exists a measurable cardinal.

Proof. Let j : V → M be a nontrivial embedding. Notice that there exists anordinal α such that j(α) �= α; otherwise, we would have rank(jx) = rank(x)for all x, and then we could prove by induction on rank that j(x) = x forall x.

Thus let κ be the least ordinal number such that j(κ) �= κ (and hencej(κ) > κ). It is clear that j(n) = n for all n and j(ω) = ω since 0, n + 1,and ω are absolute notions and j is elementary. Hence κ > ω. We shall showthat κ is a measurable cardinal.

Let D be the collection of subsets of κ defined as follows:

(17.2) X ∈ D if and only if κ ∈ j(X) (X ⊂ κ).

Since κ < j(κ), i.e., κ ∈ j(κ), we have κ ∈ D; also ∅ /∈ D because j(∅) = ∅.Using the fact that j(X ∩Y ) = j(X)∩ j(Y ) and that j(X) ⊂ j(Y ) whenever


X ⊂ Y , we see that D is a filter: If κ ∈ j(X) and κ ∈ j(Y ), then κ ∈ j(X∩Y );if X ⊂ Y and κ ∈ j(X), then κ ∈ j(Y ). Similarly, j(κ − X) = j(κ) − j(X)and thus D is an ultrafilter.

D is a nonprincipal ultrafilter: For every α < κ, we have j({α}) ={j(α)} = {α}, and so κ /∈ j({α}) and we have {α} /∈ D. We shall nowshow that D is κ-complete. Let γ < κ and let X = 〈Xα : α < γ〉 be a se-quence of subsets of κ such that κ ∈ j(Xα) for each α < γ. We shall showthat

⋂α<γ Xα ∈ D. In M (and thus in V ), j(X ) is a sequence of length j(γ)

of subsets of j(κ); for each α < γ, the j(α)th term of j(X ) is j(Xα). Sincej(α) = α for all α < γ and j(γ) = γ, it follows that j(X ) = 〈j(Xα) : α < γ〉.Hence if X =

⋂α<γ Xα, we have j(X) =

⋂α<γ j(Xα). Now it is clear that

κ ∈ j(X) and hence X ∈ D. ��

The construction of a κ-complete ultrafilter from an elementary embed-ding yields the following commutative diagram (17.3):

Lemma 17.4. Let j : V → M be a nontrivial elementary embedding, let κ bethe least ordinal moved, and let D be the ultrafilter on κ defined in (17.2). LetjD : V → Ult be the canonical embedding of V in the ultrapower UltD(V ).Then there is an elementary embedding k of Ult in M such that k(jD(a)) =j(a) for all a:

(17.3)

Ult

V Mj

kjD

�

�

��

��

��

Proof. For each [f ] ∈ Ult, let

(17.4) k([f ]) = (j(f))(κ).

(Here f is a function on κ and j(f) is a function on j(κ).)We shall first show that definition (17.4) does not depend on the choice

of f representing [f ]. If f =D g, then the set X = {α : f(α) = g(α)} is in Dand hence κ is in the set

j(X) = {α < j(κ) : (jf)(α) = (jg)(α)}.

Therefore (jf)(κ) = (jg)(κ).Next we show that k is elementary. Let ϕ(x) be a formula and let Ult �

ϕ([f ]); we shall show that M � ϕ(k([f ])). The set X = {α : ϕ(f(α))} is in Dand hence κ belongs to the set

j(X) = {α < j(κ) : M � ϕ((jf)(α))}.

Since (jf)(κ) = k([f ]), we have M � ϕ(k([f ])).


Finally, we show that k(jD(a)) = j(a) for all a. Since jD(a) = [ca], whereca is the constant function on κ with value a, we have k(jD(a)) = (j(ca))(κ).Now j(ca) is the constant function on j(κ) with value j(a) and hence(j(ca))(κ) = j(a). ��

We remark that the measure D = {X ⊂ κ : κ ∈ j(X)} defined froman elementary embedding is normal: Let f be a regressive function on someX ∈ D. Then (jf)(κ) < κ, and if γ = (jf)(κ), then f(α) = γ for almostall α.

Normality can be expressed in terms of ultrapowers:

Lemma 17.5. Let D be a nonprincipal κ-complete ultrafilter on κ. Then thefollowing are equivalent :

(i) D is normal.(ii) In the ultrapower UltD(V ),

κ = [d]

where d is the diagonal function.(iii) For every X ⊂ κ, X ∈ D if and only if κ ∈ jD(X).

Proof. (i) implies (ii): Every function f ∈∗ d is regressive, and hence repre-sents an ordinal γ < κ.

(ii) implies (iii): If X ⊂ κ, then X ∈ D if and only if d(α) ∈ X for almostall α, that is, if and only if [d] ∈ jD(X). If [d] = κ, we get X ∈ D if and onlyif κ ∈ jD(X).

(iii) implies (i): If D = {X ⊂ κ : κ ∈ jD(X)} then D is normal, by theremark preceding the lemma. ��

Let j : V → M be an elementary embedding. If X is a class defined bya formula ϕ, then j(X) is the class of the model M , defined in M by the sameformula ϕ. Note that j(X) =

⋃α∈Ord j(X ∩ Vα). In particular, M = j(V ).

Lemma 17.6. Let j be an elementary embedding of the universe and let κbe the least ordinal moved (i.e., j(κ) > κ). If C is a closed unbounded subsetof κ, then κ ∈ j(C).

Proof. Since j(α) = α for all α < κ, we have j(C) ∩ κ = C. Thus j(C) ∩ κis unbounded in κ; and because j(C) is closed (in j(V ) and hence in theuniverse), we have κ ∈ j(C). ��

A consequence of Lemma 17.6 is that the set of all regular cardinals belowa measurable cardinal κ is stationary (cf. Lemma 10.21): Let X ⊂ κ be theset of all regular cardinals below κ. Since κ is regular in M , we have κ ∈ j(X),and κ ∈ j(X ∩C) for every closed unbounded C ⊂ κ. Hence X is stationary.Similarly, as κ is Mahlo, it is Mahlo in M , and if X is now the set of allMahlo cardinals below κ, it follows that X is stationary.


More generally, if M(X) denotes the Mahlo operation

(17.5) M(X) = {α : X ∩ α is stationary in α}

where X is any class of ordinals, the above argument shows that if κ ∈ j(X)then κ ∈ M(X) (Exercise 17.7).

The next theorem shows that there exists no nontrivial elementary em-bedding of V into V . As the statement “there exists an elementary embeddingof V ” is not expressible in the language of set theory, the theorem needs to beunderstood as a theorem in the following modification of ZFC: The languagehas, in addition to ∈, a function symbol j, the axioms include Separation andReplacement Axioms for formulas that contain the symbol j, and axioms thatstate that j is an elementary embedding of V .

Theorem 17.7 (Kunen). If j : V → M is a nontrivial elementary embed-ding, then M �= V .

First we prove the following lemma:

Lemma 17.8. Let λ be an infinite cardinal such that 2λ = λℵ0 . There existsa function F : λω → λ such that whenever A is a subset of λ of size λ andγ < λ, there exists some s ∈ Aω such that F (s) = γ.

Proof. Let {(Aα, γα) : α < 2λ} be an enumeration of all pairs (A, γ) whereγ < λ and A is a subset of λ of size λ. We define, by induction on α, a sequencesα, α < 2λ, of elements of λω as follows: If α < 2λ, then since λℵ0 = 2λ > |α|,there exists an sα ∈ Aω

α such that sα �= sβ for all β < α.For each α < 2λ, we define F (sα) = γα (and let F (s) be arbitrary if s is

not one of the sα). The function F has the required property: If A ⊂ λ hassize λ and γ < λ, then (A, γ) = (Aα, γα) for some α, and then γα = F (sα).

��

Proof of Theorem 17.7. Let us assume that j is a nontrivial elementary em-bedding of V in V . Let κ = κ0 be the least ordinal moved; κ0 is measurable,and so are κ1 = j(κ0), κ2 = j(κ1), and every κn, where κn+1 = j(κn). Letλ = limn→∞ κn. Since j(〈κn : n < ω〉) = 〈j(κn) : n < ω〉 = 〈κn+1 : n < ω〉,we have j(λ) = limn→∞ j(κn) = λ. Let G = {j(α) : α < λ}; we shall use theset G and Lemma 17.8 to obtain a contradiction.

The cardinal λ is the limit of a sequence of measurable cardinals and henceis a strong limit cardinal. Since cf λ = ω, we have 2λ = λℵ0 . By Lemma 17.8there is a function F : λω → λ such that F (Aω) = λ for all A ⊂ λ of size λ.Since j is elementary, and j(ω) = ω and j(λ) = λ, the function j(F ) has thesame property. Thus, considering the set A = G, there exists s ∈ Gω suchthat (jF )(s) = κ.

Now, s is a function, s : ω → G, and hence there is a t : ω → λ suchthat s(n) = j(t(n)) for all n < ω. It follows that s = j(t). Thus we haveκ = (jF )(jt) = j(F (t)); in other words, κ = j(α) where α = F (t). However,this is impossible since j(α) = α for all α < κ, and j(κ) > κ. ��


Let us now consider ultrapowers and the corresponding elementary em-beddings jU : V → Ult. To introduce the following lemma, let us observethat if j : V → M and if κ is the least ordinal moved, then j(x) = x forevery x ∈ Vκ, and j(X) ∩ Vκ = X for every X ⊂ Vκ. Hence V M

κ+1 = Vκ+1

(and PM (κ) = P (κ)).

Lemma 17.9. Let U be a nonprincipal κ-complete ultrafilter on κ, let M =UltU (V ) and let j = jU be the canonical elementary embedding of V in M .

(i) Mκ ⊂ M , i.e., every κ-sequence 〈aα : α < κ〉 of elements of M isitself a member of M .

(ii) U /∈ M .(iii) 2κ ≤ (2κ)M < j(κ) < (2κ)+.(iv) If λ is a limit ordinal and if cf λ = κ, then j(λ) > limα→λ j(α); if

cf λ �= κ, then j(λ) = limα→λ j(α).(v) If λ > κ is a strong limit cardinal and cf λ �= κ, then j(λ) = λ.

Proof. (i) Let 〈aξ : ξ < κ〉 be a κ-sequence of elements of M . For eachξ < κ, let gξ be a function that represents aξ, and let h be a function thatrepresents κ:

[gξ] = aξ, [h] = κ.

We shall construct a function F such that [F ] = 〈aξ : ξ < κ〉. We let, for eachα < κ,

F (α) = 〈gξ : ξ < h(α)〉.

Since for each α, F (α) is an h(α)-sequence, [F ] is a κ-sequence. Let ξ < κ; wewant to show that the ξth term of [F ] is aξ. Since [h] > ξ, we have ξ < h(α)for almost all α; and for each α such that ξ < h(α), the ξth term of F (α)is gξ(α). But [cξ] = ξ and [gξ] = aξ, and we are done.

(ii) Assume that U ∈ M , and let us consider the mapping e of κκ onto j(κ)defined by e(f) = [f ]. Since κκ ∈ M and U ∈ M , the mapping e is in M .It follows that M � |j(κ)| ≤ 2κ. This is a contradiction since κ < j(κ) andj(κ) is inaccessible in M .

(iii) 2κ ≤ (2κ)M holds because PM (κ) = P (κ) and M ⊂ V ; (2κ)M is lessthan j(κ) since j(κ) is inaccessible in M ; finally, we have |j(κ)| = 2κ andhence j(κ) < (2κ)+.

(iv) If cf λ = κ, let λ = limα→κ λα and let f(α) = λα for all α < κ.Then [f ] > j(λα) for all α < κ and [f ] < j(λ). If cf λ > κ, then for everyf : κ → λ there exists α < λ such that [f ] < j(α). If cf λ = γ < κ, letλ = limν→γ λν ; for every f : κ → λ there exists (by κ-completeness) ν < γsuch that [f ] < j(λν).

(v) For every α < λ, the ordinals below α are represented by functionsf : κ → α; hence |j(α)| ≤ |ακ| < λ; by (iv) we have j(λ) = limα→λ j(α) = λ.

��


Note that in (v) it suffices to assume that cf λ �= κ and ακ < λ for allcardinals α < λ.

Let us recall (Lemma 10.18) that a measurable cardinal is weakly com-pact. We now prove a stronger result:

Theorem 17.10. Every measurable cardinal κ is weakly compact and if D isa normal measure on κ then the set {α < κ : α is weakly compact} is in D.

Proof. The first statement was proved in Lemma 10.18. Let D be a normalmeasure on κ, and let jD : V → M be the canonical embedding. SincePM (κ) = P (κ), it follows that κ is weakly compact in M , and since [d]D = κ,we have {α : α is weakly compact} ∈ D. ��

The following two results show that the existence of measurable cardinalsinfluences cardinal arithmetic:

Lemma 17.11. Let κ be a measurable cardinal. If 2κ > κ+, then the set{α < κ : 2α > α+} has measure one for every normal measure on κ.

Consequently, if 2α = α+ for all cardinals α < κ, then 2κ = κ+.

Proof. Let D be a normal measure on κ, and let M = UltD(V ). If 2α = α+

for almost all α, then, since [d]D = κ, we have M � 2κ = κ+. In other words,there is a one-to-one mapping in M between PM (κ) and (κ+)M . However,PM (κ) = P (κ) and (κ+)M = κ+ (because PM (κ) = P (κ)), and so 2κ = κ+.

��

Lemma 17.12. Let κ be a measurable cardinal, let D be a normal measureon κ and let j : V → M be the corresponding elementary embedding. Letλ > κ be a strong limit cardinal of cofinality κ. Then 2λ < j(λ).

Proof. Since cf λ = κ, we have j(λ) > λ. We shall show that 2λ = λκ ≤(λκ)M ≤ (λj(κ))M < j(λ). The first equality holds because λ is strong limit.We have λκ ≤ (λκ)M because every function f : κ → λ is in M . As for thelast inequality, we have

M � j(λ) is a strong limit cardinal

and since λ < j(λ) and j(κ) < j(λ), we have M � λj(κ) < j(λ). ��

See Exercises 17.12–17.16.

Weak Compactness

We shall investigate weakly compact cardinals in some detail, and give a char-acterization of weakly compact cardinals that explains the name “weaklycompact.” This aspect of weakly compact cardinals has, as many other largecardinal properties, motivation in model theory.


We shall consider infinitary languages which are generalizations of theordinary first order language. Let κ be an infinite cardinal number. Thelanguage Lκ,ω consists of

(i) κ variables;(ii) various relation, function, and constant symbols;(iii) logical connectives and infinitary connectives

∨ξ<α ϕξ,

∧ξ<α ϕξ for

α < κ (infinite disjunction and conjunction);(iv) quantifiers ∃v, ∀v.

The language Lκ,κ is like Lκ,ω except that it also contains infinitary quanti-fiers:

(v) ∃ξ<αvξ, ∀ξ<αvξ for α < κ.

The interpretation of the infinitary symbols of Lκ,κ is the obvious generaliza-tion of the finitary case where

∨ξ<n ϕξ is ϕ0 ∨ . . .∨ ϕn−1, ∃ξ<nvξ stands for

∃v0 . . .∃vn−1, etc. The language Lω,ω is just the language of the first orderpredicate calculus.

The finitary language Lω,ω satisfies the Compactness Theorem: If Σ isa set of sentences such that every finite S ⊂ Σ has a model, then Σ hasa model. Let us say that the language Lκ,κ (or Lκ,ω) satisfies the WeakCompactness Theorem if whenever Σ is a set of sentences of Lκ,κ (Lκ,ω) suchthat |Σ| ≤ κ and that every S ⊂ Σ with |S| < κ has a model, then Σ hasa model. Clearly, if Lκ,κ satisfies the Weak Compactness Theorem, then sodoes Lκ,ω because Lκ,ω ⊂ Lκ,κ.

Theorem 17.13.

(i) If κ is a weakly compact cardinal, then the language Lκ,κ satisfies theWeak Compactness Theorem.

(ii) If κ is an inaccessible cardinal and if Lκ,ω satisfies the Weak Com-pactness Theorem, then κ is weakly compact.

Proof. (i) The proof of the Weak Compactness Theorem for Lκ,κ is very muchlike the standard proof of the Compactness Theorem for Lω,ω. Let Σ be a setof sentences of Lκ,κ of size κ such that if S ⊂ Σ and |S| < κ, then S hasa model. Let us assume that the language L = Lκ,κ has only the symbolsthat occur in Σ; thus |L| = κ.

First we extend the language as follows: For each formula ϕ with freevariables vξ, ξ < α, we introduce new constant symbols cϕ

ξ , ξ < α (Skolemconstants); let L(1) be the extended language. Then we do the same for eachformula of L(1) and obtain L(2) ⊃ L(1). We do the same for each n < ω, andthen let L∗ =

⋃∞n=1 L(n). Since κ is inaccessible, it follows that |L∗| = κ.

L∗ has the property that for each formula ϕ with free variables vξ, ξ < α,there are in L∗ constant symbols cϕ

ξ , ξ < α (which do not occur in ϕ).


For each ϕ(vξ, . . .)ξ<α let σϕ be the sentence (a Skolem sentence)

(17.6) ∃ξ<αvξ ϕ(vξ , . . .)ξ<α → ϕ(cϕξ , . . .)ξ<α

and let Σ∗ = Σ ∪ {σϕ : ϕ is a formula of L∗}.Note that if S ⊂ Σ∗ and |S| < κ, then S has a model: Take a model for

S∩Σ (for L) and then expand it to a model for L∗ by interpreting the Skolemconstants so that each sentence (17.6) is true.

Let {σα : α < κ} be an enumeration of all the sentences in L∗. Let (T,⊂)be the binary κ-tree consisting of all t : γ → {0, 1}, γ < κ, for which thereexists a model A of Σ ∩ {σα : α ∈ dom(t)} such that for all α ∈ dom(t)

t(α) = 1 if and only if A � σα.

Since κ has the tree property, there exists a branch B in T of length κ. Let

∆ = {σα : t(α) = 1 for some t ∈ B}.

Clearly, Σ∗ ⊂ ∆. Let A0 be the set of all constant terms of L∗, and let ≈ bethe equivalence relation on A0 defined by

τ1 ≈ τ2 if and only if (τ1 ≈ τ2) ∈ ∆,

and let A = A0/≈.We make A into a model A for L∗ as follows:

A � P [[τ1], . . . , [tn]] if and only if P (τ1, . . . , τn) ∈ ∆

and similarly for function and constant symbols. The proof is then completedby showing that A is a model for ∆ (and hence for Σ). The proof of

(17.7) A � σ if and only if σ ∈ ∆

is done by induction on the number of quantifier blocks in σ: If σ =∃ξ<αvξ ϕ(vξ, . . .), then by induction hypothesis we have

A � σ(cϕξ , . . .)ξ<α if and only if σ(cϕ

ξ , . . .)ξ<α ∈ ∆

and (17.7) follows.(ii) Let κ be inaccessible and assume that the language Lκ,ω satisfies the

Weak Compactness Theorem. We shall show that κ has the tree property.Let (T, <) be a tree of height κ such that each level of T has size < κ.Let us consider the Lκ,ω language with one unary predicate B and constantsymbols cx for all x ∈ T . Let Σ be the following set of sentences:

¬(B(cx) ∧ B(cy)) for all x, y ∈ T that are incomparable,∨x∈Uα

B(cx) for all α < κ, where Uα is the αth level of T

(Σ says that B is branch in T of length κ). If S ⊂ Σ and |S| < κ, then weget a model for S by taking a sufficiently large initial segment of T and somebranch in this segment. By the Weak Compactness Theorem for Lκ,ω, Σ hasa model, which obviously yields a branch of length κ. ��


Indescribability

Let n > 0 be a natural number and let us consider the nth order predicatecalculus. There are variables of orders 1, 2, . . . , n, and the quantifiers areapplied to variables of all orders. An nth order formula contains, in additionto first order symbols and higher order quantifiers, predicates X(z) whereX and z are variables of order k + 1 and k respectively (for any k < n).Satisfaction for an nth order formula in a model A = (A, P, . . . , f, . . . , c, . . .)is defined as follows: Variables of first order are interpreted as elements ofthe set A, variables of second order as elements of P (A) (as subsets of A),etc.; variables of order n are interpreted as elements of Pn−1(A). The predi-cate X(z) is interpreted as z ∈ X . A Πn

m formula is a formula of order n + 1of the form

(17.8) ∀X ∃Y . . .︸︷︷︸m quantifiers

ψ

where X , Y , . . . are (n + 1)th order variables and ψ is such that all quantifiedvariables are of order at most n. Similarly, a Σn

m formula is as in (17.8), butwith ∃ and ∀ interchanged.

We shall often exhibit a sentence σ and claim that it is Πnm (or Σn

m)although it is only equivalent to a Πn

m (or Σnm) sentence, in the following

sense: We are considering a specific type of models in which σ is interpreted(e.g., the models (Vα,∈)) and there is a Πn

m (or Σnm) sentence σ such that

the equivalence σ ↔ σ holds in all these models.Note that every first order formula is equivalent to some Π0

n formula (andalso to some Σ0

k formula).

Definition 17.14. A cardinal κ is Πnm-indescribable if whenever U ⊂ Vκ

and σ is a Πnm sentence such that (Vκ,∈, U) � σ, then for some α < κ,

(Vα,∈, U ∩ Vα) � σ.

Lemma 17.15. Every measurable cardinal is Π21-indescribable.

Proof. Let κ be a measurable cardinal, let U ⊂ Vκ and let σ be a Π21 sentence

of the (third order) language {∈, U}. Let us assume that (Vκ,∈, U) � σ.We have σ = ∀X ϕ(X) where X is a third order variable and ϕ(X) con-

tains only second and first order quantifiers. Thus

(17.9) ∀X ⊂ Vκ+1 (Vκ+1,∈, X, Vκ, U) � ϕ

where ϕ is the (first order) sentence obtained from ϕ by replacing the firstorder quantifiers by the restricted quantifiers ∀x ∈ Vκ and ∃x ∈ Vκ.

Now let D be a normal measure on κ and let M = UltD(V ). SinceV M

κ+1 = Vκ+1, we know that (17.9) holds also in M . Using the fact that Vκ is


represented in the ultrapower by the function α �→ Vα, Vκ+1 by α �→ Vα+1,and U by α �→ U ∩ Vα, we conclude that for almost all α,

(17.10) ∀X ⊂ Vα+1 (Vα+1,∈, X, Vα, U ∩ Vα) � ϕ.

Then, translating (17.10) back into the third order language, we obtain

(Vα,∈, U ∩ Vα) � σ

for almost all, and hence for some, α < κ. ��

Lemma 17.16. If κ is not inaccessible, then it is describable by a first ordersentence, i.e., Π0

m-describable for some m.

Proof. Let κ be a singular cardinal, and let f be a function with dom(f) =λ < κ and ran(f) cofinal in κ. Let U1 = f and U2 = {λ}, and let σ bethe first order sentence saying that U2 is nonempty and that the uniqueelement of U2 is the domain of U1. Clearly, κ is describable in the sense that(Vκ,∈, U1, U2) � σ and there is no α < κ such that (Vα,∈, U1 ∩ Vα, U2 ∩ Vα) �σ. It is routine to find a single U ⊂ Vκ and an (∈, U)-sentence σ attesting tothe describability of κ.

If κ ≤ 2λ for some λ < κ, there is a function f that maps P (λ) onto κ.We let U1 = f and U2 = {P (λ)}; then κ is described by the same sentenceas above.

Finally, κ = ω is describable as follows: (Vκ,∈) � ∀x∃y x ∈ y. ��

The converse is also true; cf. Exercise 17.23.We shall now present a result of Hanf and Scott that shows that Π1

1-indescribable cardinals are exactly the weakly compact cardinals. First weneed a lemma.

Lemma 17.17. If κ is a weakly compact cardinal, then for every U ⊂ Vκ,the model (Vκ,∈, U) has a transitive elementary extension (M,∈, U ′) suchthat κ ∈ M .

Proof. Let Σ be the set of all Lκ,κ sentences true in the model (Vκ,∈, U, x)x∈Vκ

plus the sentencesc is an ordinal,

c > α, (all α < κ).

Clearly |Σ| = κ, and if S ⊂ Σ is such that |S| < κ, then S has a model(namely Vκ, where the constant c can be interpreted as some ordinal greaterthan all the α’s mentioned in S).

Hence Σ has a model A = (A, E, UA, xA)x∈Vκ ; we may assume that A ⊃Vκ, E ∩ (Vκ × Vκ) = ∈, UA ∩ Vκ = U , and xA = x for all x ∈ Vκ. Moreover,Vκ ≺ (A, E, UA) because A satisfies all formulas true in Vκ of all x ∈ Vκ.If we show that the model (A, E) is well-founded, then the lemma follows.


Here we make use of the expressive power of the infinitary language Lκ,κ: Weconsider the sentence

(17.11) ¬∃v0 ∃v1 . . . ∃vn . . .∧

n∈ω(vn+1 ∈ vn).

The sentence (17.11) holds in a model A = (A, E) if and only if A is well-founded. Since Σ contains the sentence (17.11), every model of Σ is well-founded. ��

The converse is also true; this will follow from the proof of Theorem 17.18.

Theorem 17.18 (Hanf-Scott). A cardinal κ is Π11-indescribable if and

only if it is weakly compact.

Proof. First we show that every Π11-indescribable cardinal is weakly compact.

If κ is Π11-indescribable, then by Lemma 17.16, κ is inaccessible, and it suffices

to show that κ has the tree property. In fact, by the proof of Theorem 17.13(i)it suffices to consider trees (T, <) consisting of sequences t : γ → {0, 1}, γ < κ.Let T be such a tree. For every α < κ, the model (Vα,∈, T ∩ Vα) satisfies theΣ1

1 sentence

(17.12) ∃B (B ⊂ T and B is a branch of unbounded length).

Namely, let B = {t�ξ : ξ < α} where t is any t ∈ T with domain α. Since κ isΠ1

1-indescribable, the sentence (17.12) holds in (Vκ,∈, T ) and hence T hasa branch of length κ.

To show that a weakly compact cardinal is Π11-indescribable, we use

Lemma 17.17. Let κ be weakly compact, let U ⊂ Vκ and let σ be a Π11 sen-

tence true in (Vκ,∈, U). We have σ = ∀X ϕ(X) where X is a second ordervariable and ϕ has only first order quantifiers.

Let (M,∈, U ′) be a transitive elementary extension of (Vκ,∈, U) such thatκ ∈ M . Since

(∀X ⊂ Vκ) (Vκ,∈, U) � ϕ(X)

and V Mκ = Vκ, we have

(M,∈, U ′) � (∀X ⊂ Vκ) (Vκ,∈, U ′ ∩ Vκ) � ϕ(X).

Therefore,

(M,∈, U ′) � ∃α (∀X ⊂ Vα) (Vα,∈, U ′ ∩ Vα) � ϕ(X),

and so(Vκ,∈, U) � ∃α (∀X ⊂ Vα) (Vα,∈, U ′ ∩ Vα) � ϕ(X).

Hence for some α < κ, (Vα,∈, U ∩ Vα) � σ. ��

Corollary 17.19. Every weakly compact cardinal κ is a Mahlo cardinal, andthe set of Mahlo cardinals below κ is stationary.


Proof. Let C ⊂ κ be a closed unbounded set. Since κ is inaccessible, (Vκ,∈, C)satisfies the following Π1

1 sentence:

¬∃F (F is a function from some λ < κ cofinally into κ)

and C is unbounded in κ.

By Π11-indescribability, there exists a regular α < κ such that C ∩ α is un-

bounded in α; hence α ∈ C. Thus κ is Mahlo.Being Mahlo is also expressible by a Π1

1 sentence:

∀X (if X is closed unbounded, then ∃ a regular α in X)

and so the same argument as above shows that there is a stationary set ofMahlo cardinals below κ. ��

Corollary 17.20. If κ is weakly compact and if S ⊂ κ is stationary, thenthere is a regular uncountable λ < κ such that S ∩ λ is stationary in λ.

Proof. “κ is regular” is expressible by a Π11 sentence in (Vκ,∈) and so is “κ is

uncountable.” “S is stationary” is Π11 in (Vκ,∈, S): For every C, if C is closed

unbounded, then S ∩ C �= ∅. ��

Lemma 17.21. If κ is weakly compact and if A ⊂ κ is such that A ∩ α ∈ Lfor every α < κ, then A is constructible.

Proof. Let A ⊂ κ be such that A ∩ α ∈ L for all α < κ. By Lemma 17.17there is a transitive model (M,∈, A′) $ (Vκ,∈, A) such that κ ∈ M . Considerthe sentence ∀α ∃x (x is constructible and x = A ∩ α) and let α = κ. ��

Unlike measurability, weak compactness is consistent with V = L:

Theorem 17.22. If κ is weakly compact then κ is weakly compact in L.

Proof. In L, let T = (κ, <T ) be a tree of height κ such that each level of Thas size less than κ. If κ is weakly compact then T has a branch B (in theuniverse), and by Lemma 17.21, B ∈ L. Hence κ has the tree property in L,and since κ is inaccessible, it is weakly compact in L. ��

Partitions and Models

Let us consider a model A = (A, PA, . . . , FA, . . . , cA, . . .) of a (not necessarilycountable) language L = {P, . . . , F, . . . , c, . . .}. Let κ be an infinite cardinaland let us assume that the universe A of the model A contains all ordinalsα < κ, i.e., κ ⊂ A.


Definition 17.23. A set I ⊂ κ is a set of indiscernibles for the model A iffor every n ∈ ω, and every formula ϕ(v1, . . . , vn),

A � ϕ[α1, . . . , αn] if and only if A � ϕ[β1, . . . , βn]

whenever α1 < . . . < αn and β1 < . . . < βn are two increasing sequences ofelements of I.

Lemma 17.24. Let κ be an infinite cardinal and assume that

κ → (α)<ω2λ

where α is a limit ordinal and λ is an infinite cardinal. Let L be a languageof size ≤ λ and let A be a model of L such that κ ⊂ A. Then A has a set ofindiscernibles of order-type α.

Proof. Let Φ be the set of all formulas of the language L. We considerthe function F : [κ]<ω → P (Φ) defined as follows: If x ∈ [κ]n andx = {α1, . . . , αn} where α1 < . . . < αn, then

F (x) = {ϕ(v1, . . . , vn) ∈ Φ : A � ϕ[α1, . . . , αn]}.

The function F is a partition into at most 2λ pieces and thus has a homo-geneous set I ⊂ κ of order-type α. It is now easy to verify that I is a set ofindiscernibles for A. ��

We shall see later that for a given limit ordinal α, the least κ that satisfiesκ → (α)<ω is inaccessible and satisfies κ → (α)<ω

λ for all λ < κ. Now we shallprove this for Ramsey cardinals.

Lemma 17.25. If κ → (κ)<ω and if λ < κ is a cardinal, then κ → (κ)<ωλ .

Proof. Let F : [κ]<ω → λ be a partition into λ < κ pieces. We consider thefollowing partition G of [κ]<ω into two pieces: If α1 < . . . < αk < αk+1 <. . . < α2k are elements of κ and if F ({α1, . . . , αk}) = F ({αk+1, . . . , α2k}),then we let G({α1, . . . , α2k}) = 1; for all other x ∈ [κ]<ω, we let G(x) = 0.

Now, let H ⊂ κ be a homogeneous set for G, |H | = κ. We claim thatfor each k and each x ∈ [H ]2k, G(x) = 1: This is because |H | = κ > λ, andtherefore we can find α1 < . . . < αk < αk+1 < . . . < α2k in H such thatF ({α1, . . . , αk}) = F ({αk+1, . . . , α2k}).

It follows that H is homogeneous for F : If α1 < . . . < αn and β1 < . . . <βn are two sequences in H , we choose a sequence γ1 < . . . < γn in H suchthat both αn < γ1 and βn < γ1. Then

G({α1, . . . , αn, γ1, . . . , γn}) = G({β1, . . . , βn, γ1, . . . , γn}) = 1,

and hence

F ({α1, . . . , αn}) = F ({γ1, . . . , γn}) = F ({β1, . . . , βn}). ��


Corollary 17.26. If κ is a Ramsey cardinal and if A ⊃ κ is a model ofa language of size < κ, then A has a set of indiscernibles of size κ. ��

The combinatorial methods introduced in this section will now be em-ployed to obtain a result on measurable cardinals considerably stronger thanScott’s Theorem. It will be shown that if a Ramsey cardinal exists thenV = L fails in a strong way. A more extensive theory will be developed inChapter 18.

Let us make a few observations about models with definable Skolem func-tions. Let A be a model of a language L such that A ⊃ κ and let I ⊂ κ bea set of indiscernibles for A. Let us assume that the model A has definableSkolem functions ; i.e., for every formula ϕ(u, v1, . . . , vn), where n ≥ 0, thereexists an n-ary function hϕ in A such that:

(i) hϕ is definable in A, i.e., there is a formula ψ such that

y = hϕ(x1, . . . , xn) if and only if A � ψ[y, x1, . . . , xn]

for all y, x1, . . . , xn ∈ A; and(ii) hϕ is a Skolem function for ϕ.

Let B ⊂ A be the closure of I under all functions in L and the functions hϕ

for all formulas ϕ. B is an elementary submodel of A, and in fact is thesmallest elementary submodel of A that includes the set I; we call B theSkolem hull of I and say that I generates B.

We augment the language of A by adding function symbols for all theSkolem functions hϕ and call Skolem terms the terms built from variables andconstant symbols (0-ary functions) by applications of functions in L and theSkolem functions. Since B is an elementary submodel of A, the interpretationof each Skolem term t is the same in B as in A. For every element x ∈ Bthere is a Skolem term t and indiscernibles γ1 < . . . < γn, elements of I,such that x = tA[γ1, . . . , γn] = tB[γ1, . . . , γn]. Now if ψ is a formula of theaugmented language, i.e., if ψ also contains the Skolem terms, it still does notdistinguish between the indiscernibles: If α1 < . . . < αn and β1 < . . . < βn

are two sequences in I, then ψ(α1, . . . , αn) holds (either in A or in B) if andonly if ψ(β1, . . . , βn) holds.

Theorem 17.27 (Rowbottom). If κ is a Ramsey cardinal, then the set ofall constructible reals is countable. More generally, if λ is an infinite cardinalless than κ, then |PL(λ)| = λ.

Proof. Let κ be a Ramsey cardinal and let λ < κ. Since κ is inaccessible, wehave PL(λ) ⊂ Lκ. Consider the model

A = (Lκ,∈, PL(λ), α)α≤λ.

A is a model of the language L = {∈, Q, cα}α≤λ where Q is a one-placepredicate (interpreted in A as P (λ)∩L) and cα, α ≤ λ, are constant symbols


(interpreted as ordinals less than or equal to λ). Since κ is Ramsey, thereexists a set I of size κ of indiscernibles for A.

The model A has definable Skolem functions: Since κ is inaccessible, Lκ isa model of ZFC + V = L and therefore has a definable well-ordering. Thuslet B ⊂ Lκ be the elementary submodel of A generated by the set I. Everyelement x ∈ B is expressible as x = t(γ1, . . . , γn) where t is a Skolem termand γ1 < . . . < γn are elements of I.

We shall now show that the set S = PL(λ) ∩ B has at most λ elements.Since S is the interpretation in B of the one-place predicate Q, it suffices toshow that there are at most λ elements x ∈ B such that B � Q(x).

Let t be a Skolem term. Let us consider the truth value of the formula

(17.13) t(α1, . . . , αn) = t(β1, . . . , βn)

for a sequence of indiscernibles α1 < . . . < αn < β1 < . . . < βn. Theformula (17.13) is either true for all increasing sequences in I or false forall increasing sequences in I. If (17.13) is true, then it is true for any twosequences α1 < . . . < αn, β1 < . . . < βn, in I: Pick γ1, . . . , γn bigger thanboth αn and βn and then t(α1, . . . , αn) = t(γ1, . . . , γn) = t(β1, . . . , βn). If(17.13) is false, then we choose κ increasing sequences

α01 < . . . < α0

n < α11 < . . . < α1

n < . . . < αξ1 < . . . < αξ

n < . . . (ξ < κ)

in I and then t(αξ1, . . . , α

ξn) �= t(αη

1 , . . . , αηn) whenever ξ �= η. In conclusion,

the set

(17.14) {t(α1, . . . , αn) : α1 < . . . < αn are in I}

has either one or κ elements.Now we apply this to evaluate the size of the set S. We know that |S| < κ

because S ⊂ PL(λ) ⊂ P (λ) and κ is inaccessible. If t is a Skolem termfor which the set (17.14) has size κ, then t(α1, . . . , αn) is not in S, for anyα1 < . . . < αn in I; by indiscernibility, Q(t(α1, . . . , αn)) is true or falsesimultaneously for all increasing sequences in I. Thus if t(α1, . . . , αn) ∈ S,the set (17.14) has only one element.

However, since |L| ≤ λ, there are at most λ Skolem terms. And since everyx ∈ B has the form t(α1, . . . , αn) for some Skolem term and α1 < . . . < αn

in I, it follows that |S| ≤ λ.Thus we have proved that S = QB = PL(λ)∩B has at most λ elements.

Now B ≺ Lκ and |B| = κ; hence the transitive collapse of B is Lκ and wehave an isomorphism

π : B � Lκ.

Since each α ≤ λ has a name in A, we have λ ∪ {λ} ⊂ B and so π(X) = Xfor each X ⊂ λ in B. In particular π(X) = X for all X ∈ S; and sinceQπ(B) = π(S) = S, we have

S = PL(λ) ∩ π(B) = PL(λ) ∩ Lκ = PL(λ).


This completes the proof: On the one hand, we proved that |S| ≤ λ; and onthe other hand, |PL(λ)| ≥ λ; thus |PL(λ)| = λ. ��

Every Ramsey cardinal is weakly compact. Not only is the least Ramseycardinal greater than the least weakly compact but, as we show below, thereis a hierarchy of large cardinals below each Ramsey cardinal, exceeding theleast weakly compact.

Definition 17.28. For every limit ordinal α, the Erdos cardinal ηα is theleast κ such that κ → (α)<ω .

We shall prove that each ηα, if it exists, is inaccessible, and if α < β thenηα < ηβ . Note that κ is a Ramsey cardinal if and only if κ = ηκ.

Lemma 17.29. If κ → (α)<ω, then κ → (α)<ω2ℵ0 .

Proof. Let f be a partition, f : [κ]<ω → {0, 1}ω. For each n < ω, let fn =f�[κ]n, and for each κ < ω, let fn,k : [κ]n → {0, 1} be as follows:

fn,k({α1, . . . , αn}) = h(k), where h = fn({α1, . . . , αn}).

Let π be a one-to-one correspondence between ω and ω × ω such that ifπ(m) = (n, k), then m ≥ n; for each m, let gm : [κ]m → {0, 1} be thepartition defined by

gm({α1, . . . , αm}) = fn,k({α1, . . . , αn})

where (n, k) = π(m).By the assumption, there exists H ⊂ κ of order-type α which is homo-

geneous for all gm. We claim that H is homogeneous for f . If not, thenfn({α1, . . . , αn}) �= fn({β1, . . . , βn}) for some α’s and β’s in H . Then forsome k, fn,k({α1, . . . , αn}) �= fn,k({β1, . . . , βn}), contrary to the assumptionthat H is homogeneous for gm, where π(m) = (n, k). ��

Lemma 17.30. For every κ < ηα, ηα → (α)<ωκ .

Proof. Let κ < ηα, and let f : [ηα]<ω → κ. We wish to find a homogeneousset for f of order-type α. Since κ < ηα, there exists g : [κ]<ω → {0, 1} thathas no homogeneous set of order-type α. For each n, let fn = f�[ηα]n andgn = g�[κ]n, and let A be the model (Vηα ,∈, fn, gn)n=0,1,....

By Lemmas 17.29 and 17.24, the model A has a set of indiscernibles H oforder-type α. We shall show that H is homogeneous for f . It suffices to showthat for each n, the formula

(17.15) fn({α1, . . . , αn}) = fn({β1, . . . , βn})

holds in A for any increasing sequence α1 < . . . < αn < β1 < . . . < βn

of indiscernibles: Then if α1 < . . . < αn and α′1 < . . . < α′

n are arbitrary


in H , we choose β1 < . . . < βn in H such that αn < β1 and α′n < β1, and

fn({α1, . . . , αn}) = fn({α′1, . . . , α

′n}) follows from (17.15).

Thus let us assume that the negation of (17.15) holds for any α1 < . . . <αn < β1 < . . . < βn in H . Let uξ, ξ < α, be increasing n-sequences in H suchthat the last element of uξ is less than the first element of uη whenever ξ < η.Let γξ = f(uξ) for all ξ < α, and let G = {γξ : ξ < α}. By indiscernibility,and since γ0 > γ1 > . . . > γξ > . . . is impossible, we have γ0 < γ1 < . . . <γξ < . . ..

We shall reach a contradiction by showing that G is homogeneous for g.For each k, consider the formula

(17.16) g({f(uξ1), . . . , f(uξk)}) = g({f(uν1), . . . , f(uνk

)}).

By indiscernibility, either (17.16) or its negation holds for all increasing se-quences ξ1 < . . . < ξk < ν1 < . . . < νk. The inequality cannot hold becauseg takes only two values, 0 and 1, and three sequences 〈ξ1, . . . , ξk〉 would givethree different values. Thus (17.16) holds, and the same argument as earlierin this proof shows that g is constant on [G]k. ��

Theorem 17.31. Every Erdos cardinal ηα is inaccessible, and if α < β thenηα < ηβ.

Proof. First we claim that ηα is a strong limit cardinal. If κ < ηα thenbecause 2κ �→ (α)2κ (by Lemma 9.3) and ηα → (α)2κ, we have 2κ < ηα. Weshall show that ηα is regular.

Let us assume that ηα is singular and that κ = cf ηα; let ηα = limν→κ λν .For each ν < κ, let fν : [λν ]<ω → {0, 1} be such that fν has no homoge-neous set of order-type α. For each n, let fν

n = fν�[λν ]n; let A be the model(Vηα ,∈, λν , fν

n)ν<κ,n=0,1,... Since ηα is a strong limit and κ < ηα, the model Ahas a set of indiscernibles H of order-type α.

Let ν be such that λν is greater than the least element of H . Then byindiscernibility, all elements of H are smaller than λν . Since the function fν

takes only two values, it follows that for each n, it is the equality

fνn({α1, . . . , αn}) = fν

n({β1, . . . , βn})

that holds for all increasing sequences α1 < . . . < αn < β1 < . . . < βn

in H , and not its negation. Hence H is homogeneous for fν , contrary to theassumption on fν .

Finally, let α < β be limit ordinals, and let us assume that ηα = ηβ .For each ξ < ηα, there exists a function fξ : [ξ]<ω → {0, 1} that has nohomogeneous subset of ξ of order-type α. Let us define g : [ηβ ]<ω → {0, 1}by

g({ξ1, . . . , ξn}) = fξn({ξ1, . . . , ξn−1}).Now if H is homogeneous for g, then for each ξ ∈ H , H ∩ ξ is homogeneousfor fξ. Hence the order-type of each H ∩ ξ is less than α, and therefore theorder-type of H is at most α, which is less than β. A contradiction. ��


We shall now show that the least Erdos cardinal ηω is greater than theleast weakly compact cardinal. We use the following lemma, of independentinterest:

Lemma 17.32. Let M and N be transitive models of ZFC and let j : M →N be a nontrivial elementary embedding ; let κ be the least ordinal moved. IfPM (κ) = PN (κ), then κ is a weakly compact cardinal in M .

Proof. We prove a somewhat stronger statement: κ is ineffable in M (seeExercise 17.26).

Let 〈Aα : α < κ〉 ∈ M be such that Aα ⊂ α for all α. We have j(Aα) = Aα

for all α < κ, and hence j(〈Aα : α < κ〉) = (〈Aα : α < j(κ)〉 (for some Aα,κ ≤ α < j(κ)). The set Aκ is in M and witnesses ineffability of κ in M . ��

Theorem 17.33. If ηω exists then there exists a weakly compact cardinalbelow ηω.

Proof. Let hϕ, ϕ ∈ Form , be Skolem functions for the language {∈} of settheory, and let us consider the model A = (Vηω ,∈, hA

ϕ)ϕ∈Form where foreach ϕ, hA

ϕ is a Skolem function for ϕ in (Vηω ,∈). The model A has a set ofindiscernibles I of order-type ω. Let B be the closure of I under the Skolemfunctions hA

ϕ .Let us consider some nontrivial order-preserving mapping of H into H .

Using the Skolem functions, we extend this mapping (in the unique way)to a nontrivial elementary embedding of B into B. Let M be the transitiveset isomorphic to B and let j : M → M be the corresponding nontrivialelementary embedding.

Since ηω is inaccessible, Vηω is a model of ZFC and thus M is a transitivemodel of ZFC. By Lemma 17.32 there exists a weakly compact cardinal in M ,and therefore in Vηω . ��

The next result shows that the Erdos cardinal ηω is consistent with V = L.In Chapter 18 we show that the existence of ηω1 implies V �= L.

Theorem 17.34. If κ → (ω)<ω then L � κ → (ω)<ω.

Proof. Let f be a constructible partition f : [κ]<ω → {0, 1}. We claim thatif there is an infinite homogeneous set for f , then there is one in L. Let Tbe the set of all finite increasing sequences t = 〈α0, . . . , αn−1〉 in κ such thatfor every k ≤ n, f is constant on [{α0, . . . , αn−1}]k, and let us consider thetree (T,⊃); clearly, T is constructible. We note that an infinite homogeneousset for f exists if and only if (T,⊃) is not well-founded. However, being well-founded is an absolute property for models of ZFC; and so if the tree is notwell-founded, then it is not well-founded in L, and the claim follows. ��

Let us consider models of a countable language L, with a distinguishedone-place predicate Q. A model A = (A, QA, . . .) of L has type (κ, λ) if |A| = κand |QA| = λ.


Definition 17.35. A cardinal κ > ℵ1 is a Rowbottom cardinal if for everyuncountable λ < κ, every model of type (κ, λ) has an elementary submodelof type (κ,ℵ0).

An infinite cardinal is a Jonsson cardinal if every model of size κ hasa proper elementary submodel of size κ.

Every Rowbottom cardinal is a Jonsson cardinal and the following lemma,a variation on Rowbottom’s Theorem, shows that every Ramsey cardinal isa Rowbottom cardinal.

Lemma 17.36. Let κ be a Ramsey cardinal, and let λ be an infinite cardinalless than κ. Let A = (A, . . .) be a model of a language L such that |L| ≤ λ,and let A ⊃ κ. If P ⊂ A is such that |P | < κ then A has an elementarysubmodel B = (B, . . .) such that |B| = κ and |P ∩ B| ≤ λ.

Moreover, if X ⊂ A is of size at most λ, then we can find B such thatX ⊂ B.

Moreover, if κ is a measurable and D is a normal measure on κ, then wecan find B such that B ∩ κ ∈ D.

Proof. First we add to the language L one unary predicate whose interpre-tation is the set P ; we also add constant symbols for all x ∈ X . Next we findsome Skolem functions hϕ (in (A, . . . , P, x)x∈X) for every formula ϕ, andextend the language further by adding function symbols for the functions hϕ.

Next we find a set of indiscernibles I ⊂ κ, of size κ, for the expandedmodel A′; if κ is measurable and D is a normal measure, we find I ∈ D. Welet B be the elementary submodel of A′ generated by I. As in the proof ofTheorem 17.27, one proves that if |P ∩ B| < κ then |P ∩ B| ≤ λ. ��

In Chapter 18 we show that if there exists a Jonsson cardinal then V �= L.

Exercises

17.1. Let U be a nonprincipal ultrafilter on ω. Then UltU (V ) is not well-founded.[For each k ∈ ω, let fk be a function on ω such that fk(n) = n−k for all n ≥ k.

Then f0 �∗ f1 �∗ f2 �∗ . . . is a descending ∈∗-sequence in Ult.]

17.2. If U is not σ-complete, then UltU (V ) is not well-founded.[There exists a countable partition {Xn : n = 0, 1, 2, . . .} of S such that

Xn /∈ U for all n. For each k, let fk be a function on S such that fk(x) = n− k forall x ∈ Xn.]

17.3. If Ult is well-founded, then every ordinal number α is represented by a func-tion f : S → Ord .

17.4. If U is a principal ultrafilter {X ∈ S : x0 ∈ S} then [f ] = f(x0) for each f ,and jU is the identity mapping.


17.5. Let U be a nonprincipal σ-complete ultrafilter on S and let λ be the largestcardinal such that U is λ-complete. Then jU (λ) > λ.

[Let {Xα : α < λ} be a partition of S into sets of measure 0; let f be a functionon S such that f(x) = α if x ∈ Xα. Then [f ] ≥ λ.]

17.6. If j is an elementary embedding of the universe into a transitive model M ,then M =

S

α∈Ord j(Vα).

17.7. Let j be an elementary embedding of the universe and let κ be the leastordinal moved. If X is a class of ordinals such that κ ∈ j(X), then κ ∈M(X).

17.8. If j : V → M is a nontrivial elementary embedding, if κ is the least ordinalmoved, and if λ = lim{κ, j(κ), j(j(κ)), . . .}, then there exists A ⊂ λ such thatA /∈M .

[Assuming that M contains all bounded subsets of λ, the proof of Theorem 17.7shows that G /∈M .]

17.9. If κ is measurable, then there exists a normal measure D on κ such thatUltD(V ) � κ is not measurable.

[Let D be a normal measure such that jD(κ) is the least possible ordinal; letM = UltD(V ). If κ is measurable in M , then there is a normal measure U on κsuch that U ∈ M . Since P (κ) ⊂ M , we have UltU (κ, <) ∈ M . By Lemma 17.9(iii)we have jU (κ) < ((2κ)+)M . Since ((2κ)+)M < jD(κ), we get a contradiction.]

A function f on κ is monotone if f(α) ≤ f(β) whenever α < β.

17.10. Let U be a nonprincipal κ-complete ultrafilter on κ. Then U extends theclosed unbounded filter if and only if the diagonal function is the least nonconstantmonotone function in UltU .

[If U extends the closed unbounded filter and if f is monotone and regressiveon some X ∈ U , then since X is stationary, f is constant on an unbounded setand hence constant almost everywhere. If U does not extend the closed unboundedfilter, then f(α) = sup(C ∩α) (where C /∈ U is closed unbounded) is a nonconstantmonotone function regressive on X ∈ U .]

17.11. Let U be a κ-complete ultrafilter on κ, and let h : κ → κ. If D = h∗(U),then the mapping k : UltD(V ) → UltU (V ) defined by k([f ]D) = [f ◦ h]U is anelementary embedding.

17.12. If D is a normal measure on κ and {α : 2α ≤ α++} ∈ D, then 2κ ≤ κ++.More generally, if β < κ and {ℵα : 2ℵα ≤ ℵα+β} ∈ D, then 2ℵκ ≤ ℵκ+β.

[If f is such that f(ℵα) = ℵα+β for all α < κ, then [f ]D = (ℵκ+j(β))M ≤ ℵκ+β.]

17.13. If D is a normal measure on κ and {α : 2ℵα < ℵα+α} ∈ D, then 2ℵκ < ℵκ+κ.[If f(α) = ℵα+α, then [f ] = (ℵκ+κ)M .]

17.14. Let κ be measurable and let λ = ℵκ+κ be strong limit. Then 2λ < ℵ(2κ)+ .[j(λ) = (ℵj(κ+κ))

M ≤ ℵj(κ)+j(κ); j(κ) + j(κ) < (2κ)+.]

17.15. Let κ be measurable, let λ be strong limit, cf λ = κ, such that λ < ℵλ.Then 2λ < ℵλ.

[If λ = ℵα, then j(λ) = (ℵj(α))M ≤ ℵj(α), and j(α) < (ακ)+ < λ.]

17.16. Let Φ(α) denote the αth fixed point of ℵ, i.e., the αth ordinal ξ suchthat ℵξ = ξ. Let κ be measurable and let λ = Φ(κ + κ) be strong limit. Then2λ < Φ((2κ)+).

[Use the fact that (Φ(α))M ≤ Φ(α) for all α.]


17.17. If κ = λ+ is a successor cardinal, then the Weak Compactness Theoremfor Lκ,ω is false.

[Consider constants cα, α ≤ κ, a binary relation < and a ternary relation R.Consider the sentences saying that (a) < is a linear ordering; (b) cα < cβ for α < β;(c) each fx is a function, where fx(y) = z stands for R(x, y, z). Let Σ consist of thesesentences, the sentence z < x → ∃y R(x, y, z) (saying that ran(fx) ⊃ {z : z < x}),and the infinitary sentence R(x, y, z)→ W

ξ<λ(y = cξ) (saying that dom(fx) ⊂ {cξ :ξ < λ}). Show that each S ⊂ Σ, |S| ≤ λ, has a model, but Σ does not.]

17.18. If κ is a singular cardinal, then the Weak Compactness Theorem for Lκ,ω

is false.[Let A ⊂ κ be a cofinal subset of size < κ. Consider constants cα, α ≤ κ, and

a linear ordering <. There is Σ that says on the one hand that {cα : α ∈ A} is cofinalin the universe, and on the other hand that for each α < κ, if (∀β < α) cκ > cβ

then cκ > cα; and each S ⊂ Σ, |S| < κ, has a model.]

17.19. If κ is weakly compact and if (B,⊂) is a κ-complete algebra of subsetsof κ such that |B| = κ, then every κ-complete filter F on B can be extended toa κ-complete ultrafilter on B.

[Consider constants cX for all X ∈ B, and a unary predicate U . Let Σ be thefollowing set of Lκ,ω-sentences: ¬U(c∅), U(cX ) ∨ U(cκ−X) for all X ∈ B, U(cX)→U(cY ) for all X ⊂ Y in B, U(cX) for all X ∈ F , and

V

X∈A U(cX)→ U(cT A) forall A ⊂ B such that |A| < κ. Show that Σ has a model.]

17.20. If κ is inaccessible and if every κ-complete filter on any κ-complete algebra Bof subsets of κ such that |B| = κ can be extended to a κ-complete ultrafilter, thenκ is weakly compact.

[As in Lemma 10.18.]

17.21. If (P, <) is a linearly ordered set of size κ, and κ is weakly compact, thenthere is a subset W ⊂ P of size κ that is either well-ordered or conversely well-ordered by <.

17.22. The least measurable cardinal is Σ21-describable.

[∃U (U is κ-complete nonprincipal ultrafilter on κ).]

17.23. Every inaccessible cardinal is Π0m-indescribable for all m.

[Let U ⊂ Vκ. The model (Vκ,∈, U) has a countable elementary submodel M0.Let α0 < κ be such that M0 ⊂ Vα0 . For each n, let Mn+1 be an elementary submodelof (Vκ,∈, U) such that Vαn ⊂ Mn+1, and let Mn+1 ⊂ Vαn+1 . Let α = limn→ω αn;then Vα is an elementary submodel of (Vκ,∈, U).]

17.24. If κ is weakly compact, then there is no countably generated completeBoolean algebra B such that |B| = κ.

[Assume that B is such. Note that sat(B) = κ. We may assume that B =(κ, +, ·,−); let A ⊂ κ be a countable set of generators. Let U1 be the set of allpairs (u, x) such that u ∈ κ, x ⊂ κ, |x| < κ, and u =

P

x, let U2 = {A}. Letσ be the conjunction of these sentences: (a) B is a Boolean algebra and B ⊃ A(first order), (b) ∀x∃u (if x ⊂ κ, then u =

P

x) (first order), and (c) ∀X (if X ⊂ κand X is a partition of B, then ∃x (x = X)) (here x is a first order variable; thesentence (c) is Π1

1). Since (Vκ,∈, U1, U2) satisfies σ, there is some α < κ suchthat (Vα,∈, U1 ∩ Vα, U2 ∩ Vα) � σ. Then (α, +, ·,−) is a complete subalgebra of Bcontaining A.]


17.25. Let κ be a measurable cardinal. If 〈Aα : α < κ〉 is a sequence of sets suchthat Aα ⊂ α for all α < κ, then there exists an A ⊂ κ such that {α ∈ κ : A∩α = Aα}is stationary.

A cardinal κ with the property from Exercise 17.25 is called ineffable.

17.26. Let κ be ineffable and let f : [κ]2 → {0, 1} be a partition. Then there existsa stationary homogeneous set. (Hence κ is weakly compact.)

[For each α < κ let Aα = {ξ < α : f({ξ, α}) = 1}, and let A ⊂ κ be such thatS = {α : A ∩ α = Aα} is stationary. Either S ∩ A or S − A is stationary, and ishomogeneous.]

17.27. If κ is ineffable then κ is ineffable in L.[Use Lemma 17.21.]

17.28. If κ is Ramsey then ℵ1 is inaccessible in L.[Show that P L[x](ω) is countable for every x ⊂ ω.]

17.29. If M is a transitive model of ZFC and if j : M → M is a nontrivialelementary embedding, then the least ordinal κ moved by j is Πn

m-indescribablein M , for all n and m.

[If U ⊂ V Mκ , then U = j(U)∩V M

κ . If σ is a Πnm sentence and M � ((Vκ,∈, U) �

σ), then M � ((∃α < j(κ)) (Vα,∈, j(U) ∩ Vα) � σ).]

17.30. The cardinal ηω is not weakly compact.[ηω is Π1

1-describable.]

17.31. An infinite cardinal κ is a Jonsson cardinal if and only if for every F :[κ]<ω → κ there exists a set H ⊂ κ of size κ such that the image of [H ]<ω under Fis not the whole set κ.

[To show that the condition is necessary, consider the model (κ, <, F1, F2, . . .)where Fn = F �[κ]n. To show that the condition is sufficient, let A = (κ, . . .) bea model. Let {hn : n < ω} be a set of Skolem functions for A, closed undercomposition and arranged so that each hα is n-ary. For each x ∈ [κ]n, let F (x) =hn(x). If H ⊂ κ, then the image of [H ]<ω under F is an elementary submodel of A.]

17.32. ℵ0 is not a Jonsson cardinal.[Let A = (ω, f) where f(n) = n− 1 for all n > 0.]

17.33. If κ is a Rowbottom cardinal, then either κ is weakly inaccessible orcf κ = ω.

[To show that κ = λ+ is not Rowbottom, let fα be a one-to-one mapping of αonto λ, for each α, such that λ ≤ α < κ. Let A = (κ, λ, <, R) where R(α, β, γ)if and only if fα(β) = γ. If (B,B ∩ λ, <, R ∩ B3) ≺ A and |B| = κ, let α be theλth element of B; then fα(B ∩ α) ⊂ B ∩ λ and hence |B ∩ λ| = λ > ℵ0.

To show that κ is not Rowbottom if κ > cf κ = λ > ℵ0, let f be a nondecreasingfunction of κ onto λ and use f to produce a counterexample.]

Historical Notes

In [1963/64] Keisler and Tarski introduced the method of ultraproducts in thestudy of measurable cardinals, and it was established that the least measurable


cardinal is greater than the least inaccessible cardinal. Scott used the method of ul-trapowers to prove that existence of measurable cardinals contradicts the Axiom ofConstructibility. Rowbottom and Silver initiated applications of infinitary combina-torics developed by Erdos and his collaborators. Scott’s Theorem appeared in [1961]and Kunen’s Theorem in [1971a]. (Lemma 17.8 is due to Erdos and Hajnal [1966].)

In [1963/64a] Hanf studied compactness of infinitary languages; his work let tothe systematic study of Keisler and Tarski. Hanf proved that the least inaccessiblecardinal is not measurable (in fact not weakly compact); Erdos and Hajnal thenpointed out (cf. [1962]) that the same result can be proved using infinitary combi-natorics. Keisler and Tarski introduced the Mahlo operation and showed that theleast measurable cardinal is much greater than, e.g., the least Mahlo cardinal, etc.

The equivalence of various formulations of weak compactness is a result ofseveral papers. In [1963/64a] Hanf initiated investigations of compactness of in-finitary languages. Erdos and Tarski listed in [1961] several properties that weresubsequently shown mutually equivalent (for inaccessible cardinals) and provedseveral implications. These properties included the partition property κ → (κ)22,the tree property, and several other properties. Hanf and Scott [1961] introducedΠn

m-indescribability and announced Theorem 17.18. Further contributions weremade in the papers Hanf [1963/64b], Hajnal [1964], Keisler [1962], Monk andScott [1963/64], Tarski [1962], and Keisler and Tarski [1963/64]. A complete list ofequivalent formulations with the proofs appeared in Silver [1971b]. Theorem 17.22is due to Silver [1971b]. Rowbottom’s Theorem (as well as Lemma 17.36) are dueto Rowbottom [1971].

The main results on Erdos cardinals are due to Rowbottom, Reinhardt, andSilver. Rowbottom proved that if ηω1 exists, then there are only countably manyconstructible reals (see [1971]); Theorem 17.33 is due to Reinhardt and Silver [1965],and Theorem 17.34 is due to Silver [1970a].

Exercise 17.10: Ketonen [1973].Ineffable cardinals were introduced by Jensen; Exercises 17.26 and 17.27 are

due to Kunen and Jensen.Exercise 17.29: Reinhardt and Silver [1965].

Documents

17. Large Cardinals - Delft University of Technology17. Large Cardinals 287 Now let κ be a measurable cardinal, and let U be a nonprincipal κ-com- plete ultraﬁlter on κ.Letd (the